Đáp án:

$\begin{array}{l}
a)Do:\left\{ \begin{array}{l}
\left| {{x^2} - 1} \right| \ge 0\\
\left| {x + 1} \right| \ge 0
\end{array} \right. \Leftrightarrow \left| {{x^2} - 1} \right| + \left| {x + 1} \right| \ge 0\\
Khi:\left| {{x^2} - 1} \right| + \left| {x + 1} \right| = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left| {{x^2} - 1} \right| = 0\\
\left| {x + 1} \right| = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} = 1\\
x =  - 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 1/x =  - 1\\
x =  - 1
\end{array} \right.\\
 \Leftrightarrow x =  - 1\\
Vậy\,x =  - 1\\
b)Do:\left\{ \begin{array}{l}
\sqrt {{x^2} - 8x + 16}  \ge 0\\
\left| {x + 2} \right| \ge 0
\end{array} \right.\\
 \Leftrightarrow \sqrt {{x^2} - 8x + 16}  + \left| {x + 2} \right| \ge 0\\
Khi:\sqrt {{x^2} - 8x + 16}  + \left| {x + 2} \right| = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{x^2} - 8x + 16}  = 0\\
\left| {x + 2} \right| = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{{\left( {x - 4} \right)}^2}}  = 0\\
x =  - 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
x =  - 2
\end{array} \right.\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
c)Dkxd:\left\{ \begin{array}{l}
1 - {x^2} \ge 0\\
x + 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} \le 1\\
x \ge  - 1
\end{array} \right. \Leftrightarrow  - 1 \le x \le 1\\
Do:\left\{ \begin{array}{l}
\sqrt {1 - {x^2}}  \ge 0\\
\sqrt {x + 1}  \ge 0
\end{array} \right.\\
 \Leftrightarrow Khi:\sqrt {1 - {x^2}}  + \sqrt {x + 1}  = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
1 - {x^2} = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} = 1\\
x =  - 1
\end{array} \right. \Leftrightarrow x =  - 1\\
Vậy\,x =  - 1\\
d)Dk:{x^2} - 4 \ge 0 \Leftrightarrow {x^2} \ge 4\\
DO:\left\{ \begin{array}{l}
\sqrt {{x^2} - 4}  \ge 0\\
\sqrt {{x^2} + 4x + 4}  \ge 0
\end{array} \right.\\
 + Khi:\sqrt {{x^2} - 4}  + \sqrt {{x^2} + 4x + 4}  = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{x^2} - 4}  = 0\\
\sqrt {{x^2} + 4x + 4}  = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} = 4\\
\sqrt {{{\left( {x + 2} \right)}^2}}  = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 2/x = 2\\
x =  - 2
\end{array} \right.\\
 \Leftrightarrow x =  - 2\\
Vậy\,x =  - 2
\end{array}$